230. Kth Smallest Element in a BST

题目

Given a binary search tree, write a function 

kthSmallest

 to find the kth
smallest element in it.

Note: 

You may assume k is always valid, 1 ≤ k ≤ BST's total elements.

Follow up:

What if the BST is modified (insert/delete operations) often and you need to find the kth smallest frequently? How would you optimize the kthSmallest routine?

Hint:
Try to utilize the property of a BST.
What if you could modify the BST node's structure?
The optimal runtime complexity is O(height of BST).
分析
在二叉检索树中查找第k小的元素，由于二叉检索树的左子树是较小元素存放位置，故使用中根遍历，每遍历一个元素k值减1，当k为0时即找到第k小的元素。

/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int kthSmallest(TreeNode* root, int k) {
if(root==NULL||k<1)
return 0;
stack<TreeNode*> s;//使用迭代法遍历二叉树
s.push(root);
TreeNode* r=root->left;//指向左子树
while(r||!s.empty())//设置r作为循环条件防止在[1,null,2]中出现1从栈中弹出，r指向2，但是在未压入栈前栈为空，无法进入循环的情况
{
while(r)//沿左子树向下查找
{
s.push(r);
r=r->left;
}
r=s.top();//遍历根节点
cout<<r->val<<endl;
s.pop();
k--;
if(k==0)
return r->val;
r=r->right;//遍历右子树
}
return 1;
}
};